A quadratic equation is an equation whose power is in the second degree. In an equation, at least one of the variables is squared. In finding the missing value, it can be solved by factoring, completing the square and using the quadratic formula.

**Example:
**\( x^2\) + 2x +1 = 0

The first term \(x^2\) is in the second degree or second power.

**Solve Quadratic Equations by Factoring
**To factorise the quadratic equation:

a. Find two numbers whose sum are of the second term and a product of the third term.

b. Extract the value of x.

c. Check by substituting each value of x in the equation.

**Example: **\(x^2\) + 5x +6 = 0

\(x^2\) + 5x +6 = 0 Think of two number that add up to 5 and are the product of 6.

The possibilities are 2 and 3.

(x + 2) (x + 3) = 0 Find the value of x.

(x + 2) = 0 (x + 3) = 0

x = -2 x = -3

Check by substituting each value of x to \(x^2\) + 5x +6 = 0

\(x^2\) + 5x +6 = 0

x = -2

\({(-2)}^2\) + 5(-2) +6 = 0

4 – 10 + 6 = 0

0 = 0

x = -3

({(-3)}^2\) + 5(-3) +6 = 0

9 – 15 + 6 = 0

0 = 0

**
Solve Quadratic Equations by Completing the Square
**Completing the square is used if the equation cannot be factorise. It is written as square plus another term. In a standard form of a quadratic equation,

\(ax^2\) + bx +c = 0 to \({a(x + d)}^2\) + e = 0 where \(d \ =\ \frac{b}{2a} \ and \ e \ = \ c – \frac{b^2}{4a}\)

*Note: *Completing the square is manipulating the equation to make sure the left is a perfect square.

**Example:** \(x^2\) + 4x +5 = 0

Get the value of d and e if a = 1, b = 4, c = 5.

\( d = \frac{b}{2a}\)

\( d = \frac{4}{2 (1)}= 2\)

\( e \ = \ c-\frac{b^2}{4a}\)

\(e = 5 -\frac{4^2}{4(1)} = 5- \frac{16}{4} = 5 – 4 = 1\)

Therefore, the equation will be \({(x + 2)}^2\) + 1 = 0.

**Solving Using the Quadratic Formula
**The best and easiest way to find the roots of the quadratic equation is by factoring. If factoring is not possible, use the quadratic formula. Quadratic formula can be used for all quadratic equations that can be solved. In the standard way of writing the quadratic equation,

\(ax^2\) + bx +c = 0, the quadratic formula is \(x=\frac{{-b}±\sqrt{{b}^2-4ac}}{2a}\).

*Note:* The value of x can be obtained by direct substitution in the given quadratic formula.

**Example 1: ** \(x^2\) + 5x +6 = 0

a = 1, b = 5, c = 6

\(x=\frac{{-b}±\sqrt{{b}^2-4ac}}{2a}\) Substitute the value of a, b and c

\(x=\frac{{-5}±\sqrt{{5}^2-4(1)(6)}}{2(1)}\)

\(x=\frac{{-5}±\sqrt{25-24}}{2}\)

\(x=\frac{{-5}±\sqrt{1}}{2}\)

\(x=\frac{{-5}±{1}}{2}\)

\(x= \frac{{-5 }+{ 1}}{2}\) \(x = \frac{{-5} -{1}}{2}\)

\( x = -2\) \( x = -3\)

**Example 2:**

\(3x^2\) – 11x +7 = 0

a = 3, b = – 11, c = 7

\(x=\frac{{-b}±\sqrt{{b}^2-4ac}}{2a}\) Substitute the value of a, b and c

\(x=\frac{-{(-11)}±\sqrt{{(-11)}^2-4(3)(7)}}{2(3)}\)

\(x=\frac{{11}±\sqrt{37}}{6}\)

\(x=\frac{{11}±{6.0828}}{6}\)

\(x = \frac{{11}+{6.0828}}{6}\) \(x = \frac{{11}-{6.0828}}{6}\)

\(x =2.84\) \( x =0.819\)

The solution set is 2.84 and 0.819.

*Note:* If the answer in b2- 4ac is negative, there are no real solutions since there is no root for a negative number.